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Hausdorff and Hausdorff and Non-Hausdorff Non-Hausdorff Spaces Spaces Mayra Ibarra Mayra Ibarra MATH101 MATH101
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Hausdorff and Non-Hausdorff Spaces

### Transcript of Hausdorff and Non-Hausdorff Spaces

• 1. Hausdorff and Non-Hausdorff Spaces Mayra Ibarra MATH101

2. What is a Topological Space?

• Recall:
• A topological space is an ordered pair (X, ), where X is a set and T is a collection of subsets of X such that
• and X
• U V whenever U, V ;
• { U a: a I } whenever {U a: a I } . 1

3. Separation Axioms 2

• T 0: If a, b X, there exists an open set O such that either a O and bO, or b O and aO.
• T 1: If a, b X, there exist open sets O a , O b containing a and b respectively, such that bO aand aO b .

4. T 2: If a, b X, there exist disjoint open sets O aand O bcontaining a and b respectively. 5. 6. T 3: If A is a closed set and b is a point not in A, there exist disjoint open sets O Aand O bcontaining A and b respectively.X 7. T 4: If A and B are disjoint closed sets in X, there exist disjoint open sets O Aand O Bcontaining A and B respectively X 8. T 5: If A and B are separated sets in X, there exist disjoint open sets O Aand O Bcontaining A and B respectively 9. Hausdorff Spaces

• All metric spaces are Hausdorff 3
• The real number line with the usual topology is Hausdorff 4
• Suppose we have an open set, S, of real numbers
• For eachp S we can find an >0, such that the -neighborhood ofpforms an open set inside of S.
• Supposep, q S and are separated by a distance r.We can pick 1and 2forpandq , respectively, such that r/2 1> 0 and r/2 2> 0.Thuspandqwill be separated by disjoint open sets.

10. Subspaces of Hausdorff Spaces are also Hausdorff 5

• If Y X and (X, d) is Hausdorff, then any two distinct points x 1 , x 2Y will also be in X. Since x 1 and x 2are in X, there are disjoint open sets O 1and O 2containing x 1 and x 2 , respectively. Thus Y contains points that can be separated by disjoint open sets and is Hausdorff.

11. Non-Hausdorff Spaces

• Zariski Topology
• Let X be [0, 1][2, 3]. Define the following equivalence relation on X:
• a ~ a + 2 for all 0 a 1 except for a =
• b ~ b 2for 2 b 3 except for b = 5/2

  5/2 1/2 12. Proof:

• Notice that every point in [0, 1] is identified with a point of [2, 3] except fpr the two points and 5/2, which remain unidentified. Thus are contained in X/~. A set in X/~ is open iff it is open in X. However, we see that there are no disjoint open sets that contain and 5/2. So X is Non-Hausdorff when considered under this equivalence relation.6

13. Regular and Normal Spaces 7

• Regular Space 8-A space which is both a T 1and a T 3space

Normal Space 9 -A space which is both a T 1and a T 4space 14.

• Every Normal Space is Regular
• Every Regular Space is Hausdorff
• Not all Hausdorff Spaces are Regular
• In the definitions of Normal and Regular we can replace the use of disjoint open sets, and use disjoint closures to yield the same spaces
• The same is not true for the definition of Hausdorff

15. Completely Hausdorff 10

• T 2 Axiom- If a and b are two points of a topological space X, there exist open sets O aand O bcontaining a and b, respectively, such that a b=
• A space that satisfies this axiom is called a completely Hausdorff space.

16. 17. Compact Hausdorff Spaces

• If a Hausdorff space is compact, then it is normal, and therefore also regular.

18.

• Let the sequence (S n ) in a Hausdorff Space have a limit, s. Suppose that the sequence has another limit, p, such that ps. Let U and V be disjoint neighborhoods of s and p, respectively. Then if we choose a sufficiently large n, S n U however, for a sufficiently large n, S n V. This contradicts the Hausdorff property because U V . Thus the sequence cannot have more than one limit.

19. If X is Hausdorff and (S n ) is a sequence in X that converges to a point s X, and if y is an accumulation point of the set{S n|n = 1, 2, . . .}, then s = y. 20.

• Suppose sy. Then there exist open sets U, VXfor s and y respectively such that UV = . Also, since (S n ) converges to s, there exists a natural number, N such that n>N implies that S nU.
• Letibe such that1
• Define V i = V iin case ys i, and V i = V in case y=s i and define V as the intersection of V with a finite collection of open sets, V i . So yVa system of neighborhoods around y.

21.

• Let z be such that z V and yz, then zs ifor anyisince ifi> N, s i U and UV UV = ,
• And ifi N, then either y=s iz, or s i W i, and W iV W i V i= .
• Thus there is a neighborhood V of y, such that no point z y of the set {S n| n = 1, 2, . . .}, belongs to V.
• This contradicts the fact that y is an accumulation point of the set {S n| n = 1, 2, . . .},
• thus s=y.

22. Endnotes

• Greever, John.Theory and Examples of Point-Set Topology.Claremont: Waybread Publications, 1990.
• Steen, Lynn Arthur and J. Arthur Seebach, Jr.Counterexamples in Topology.New York: Dover Publications Inc., 1995.
• Image of axiom spaces: http://jtauber.com/2005/01/separation.png
• Sneddon, I. N. Ed. Andrew H. Wallace. An Introduction To Algebraic Topology.International Series of Monographs in Pure and Applied Mathematics.V.I. New York: Pergamon Press, 1957.
• Baum, John D.Elements of Point Set Topology . New York: Dover Publications Inc., 1991
• Sneddon, pg. 32.
• Goodman, Sue E.Beginning Topology.Belmont: Brooks/Cole, 2005.
• Baum, pg. 81.
• Image-http://en.wikipedia.org/wiki/Regular_space
• Image-http://commons.wikimedia.org/wiki/Image:Normal_space.svg
• Steen, pg. 13.
• Steen, pg. 13.
• Sneddon, pg. 31.